summaryrefslogtreecommitdiff
path: root/macros/latex/contrib/tkz/tkz-euclide/doc/latex/TKZdoc-euclide-triangles.tex
blob: 990862afff2059697a38998798edaa04181310b3 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
\section{Triangles}

\subsection{Definition of triangles \tkzcname{tkzDefTriangle}}
The following macros will allow you to define or construct a triangle from \tkzname{at least} two points.

 At the moment, it is possible to define the following triangles:
 \begin{itemize}
\item  \tkzname{two angles}  determines a triangle with two angles;
\item  \tkzname{equilateral}  determines an equilateral triangle;
\item  \tkzname{isosceles right}  determines an isoxsceles right triangle;
\item \tkzname{half} determines a right-angled triangle such that the ratio of the measurements of the two adjacent sides to the right angle is equal to $2$;
\item \tkzname{pythagore} determines a right-angled triangle whose side measurements are proportional to 3, 4 and 5;
\item \tkzname{school} determines a right-angled triangle whose angles are 30, 60 and 90 degrees;
\item \tkzname{golden} determines a right-angled triangle such that the ratio of the measurements on the two adjacent sides to the right angle is equal to $\Phi=1.618034$, I chose "golden triangle" as the denomination because it comes from the golden rectangle and I kept the denomination "gold triangle" or "Euclid's triangle" for the isosceles triangle whose angles at the base are 72 degrees;

\item  \tkzname{euclid} or \tkzname{gold} for the gold triangle; in the previous version the option was "euclide" with an "e".

\item \tkzname{cheops} determines a third point such that the triangle is isosceles with side measurements proportional to $2$, $\Phi$ and $\Phi$.
\end{itemize}

\begin{NewMacroBox}{tkzDefTriangle}{\oarg{local options}\parg{A,B}}%
The points are ordered because the triangle is constructed following the direct direction of the trigonometric circle. This macro is either used in partnership with \tkzcname{tkzGetPoint} or by using \tkzname{tkzPointResult} if it is not necessary to keep the name.

\medskip
\begin{tabular}{lll}%
\toprule
options             & default & definition                        \\
\midrule
\TOline{two angles= \#1 and \#2}{no defaut}{triangle knowing two angles}
\TOline{equilateral} {equilateral}{equilateral triangle }
\TOline{half} {equilateral}{B rectangle  $AB=2BC$ $AC$ hypothenuse }
\TOline{isosceles right} {equilateral}{isosceles right triangle }
\TOline{pythagore}{equilateral}{proportional to the pythagorean triangle 3-4-5}
\TOline{pythagoras}{equilateral}{same as above}
\TOline{egyptian}{equilateral}{same as above}
\TOline{school} {equilateral}{angles of 30, 60 and 90 degrees }
\TOline{gold}{equilateral}{angles of 72, 72 and 36 degrees, $A$ is the apex}
\TOline{euclid} {equilateral}{same as above but $[AB]$ is the base}
\TOline{golden} {equilateral}{B rectangle and $AB/AC = \Phi$}
\TOline{cheops} {equilateral}{AC=BC, AC and BC are proportional to $2$ and $\Phi$.}
\TOline{swap} {false}{gives the symmetric point with respect to $AB$}
\bottomrule
\end{tabular}

\medskip
\tkzcname{tkzGetPoint} allows you to store the point otherwise \tkzname{tkzPointResult} allows for immediate use.
\end{NewMacroBox}

\subsubsection{Option \tkzname{equilateral}}
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(4,0){B}
  \tkzDefTriangle[equilateral](A,B)
  \tkzGetPoint{C}
  \tkzDrawPolygons(A,B,C)
  \tkzDefTriangle[equilateral](B,A)
  \tkzGetPoint{D}
  \tkzDrawPolygon(B,A,D)
  \tkzMarkSegments[mark=s|](A,B B,C A,C A,D B,D)
\end{tikzpicture}
\end{tkzexample}


\subsubsection{Option \tkzname{two angles}}
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A} 
\tkzDefPoint(5,0){B} 
\tkzDefTriangle[two angles = 50 and 70](A,B)
\tkzGetPoint{C} 
\tkzDrawSegment(A,B) 
\tkzDrawPoints(A,B) 
\tkzLabelPoints(A,B) 
\tkzDrawSegments[new](A,C B,C) 
\tkzDrawPoints[new](C)
\tkzLabelPoints[above,new](C)
\tkzLabelAngle[pos=1.4](B,A,C){$50^\circ$}
\tkzLabelAngle[pos=0.8](C,B,A){$70^\circ$}
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{school}}
The angles are 30, 60 and 90 degrees.

\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}
  \tkzDefPoints{0/0/A,4/0/B}
  \tkzDefTriangle[school](A,B)  
  \tkzGetPoint{C}
  \tkzMarkRightAngles(C,B,A)
  \tkzLabelAngle[pos=0.8](B,A,C){$30^\circ$}
  \tkzLabelAngle[pos=0.8](C,B,A){$90^\circ$}
  \tkzLabelAngle[pos=0.8](A,C,B){$60^\circ$} 
  \tkzDrawSegments(A,B)
  \tkzDrawSegments[new](A,C B,C)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{pythagore}}
This triangle has sides whose lengths are proportional to 3, 4 and 5.

\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture} 
  \tkzDefPoints{0/0/A,4/0/B} 
  \tkzDefTriangle[pythagore](A,B) 
  \tkzGetPoint{C} 
  \tkzDrawSegments(A,B)
  \tkzDrawSegments[new](A,C B,C)
  \tkzMarkRightAngles(A,B,C)
  \tkzLabelPoint[above,new](C){$C$} 
  \tkzDrawPoints[new](C) 
  \tkzDrawPoints(A,B) 
  \tkzLabelPoints(A,B) 
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{pythagore} and \tkzname{swap}}
This triangle has sides whose lengths are proportional to 3, 4 and 5.

\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture} 
  \tkzDefPoints{0/0/A,4/0/B} 
  \tkzDefTriangle[pythagore,swap](A,B) 
  \tkzGetPoint{C} 
  \tkzDrawSegments(A,B)
  \tkzDrawSegments[new](A,C B,C)
  \tkzMarkRightAngles(A,B,C)
  \tkzLabelPoint[above,new](C){$C$} 
  \tkzDrawPoints[new](C) 
  \tkzDrawPoints(A,B) 
  \tkzLabelPoints(A,B) 
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{golden}}
\begin{tkzexample}[latex=6 cm,small]
\begin{tikzpicture}[scale=.8]
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B} 
\tkzDefTriangle[golden](A,B)\tkzGetPoint{C} 
\tkzDefSpcTriangle[in,name=M](A,B,C){a,b,c}
\tkzDrawPolygon(A,B,C) 
\tkzDrawPoints(A,B) 
\tkzDrawSegment(C,Mc) 
\tkzDrawPoints[new](C)
\tkzLabelPoints(A,B) 
\tkzLabelPoints[above,new](C)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{isosceles right}}
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(4,0){B}
  \tkzDefTriangle[isosceles right](A,B)
  \tkzGetPoint{C}
  \tkzDrawPolygons(A,B,C)
  \tkzDrawPoints(A,B,C)
  \tkzMarkRightAngles(A,C,B)
  \tkzLabelPoints(A,B,C)
\end{tikzpicture}
\end{tkzexample}

\subsubsection{Option \tkzname{gold} }
\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}
  \tkzDefPoints{0/0/A,4/0/B} 
  \tkzDefTriangle[gold](A,B)
  \tkzGetPoint{C}
  \tkzDrawPolygon(A,B,C)
  \tkzDrawPoints(A,B,C)
  \tkzLabelPoints(A,B)
  \tkzLabelPoints[above](C)
 \tkzLabelAngle[pos=0.8](B,A,C){$36^\circ$}
 \tkzLabelAngle[pos=0.8](C,B,A){$72^\circ$}
 \tkzLabelAngle[pos=0.8](A,C,B){$72^\circ$}
\end{tikzpicture}
\end{tkzexample}


\subsubsection{Option \tkzname{euclid}}
\tkzimp{Euclid} and \tkzimp{gold} are identical but the segment AB is a base in one and a side in the other. 

\begin{tkzexample}[latex=7 cm,small]
\begin{tikzpicture}[scale=.75]
 \tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
 \tkzDefTriangle[euclid](A,B)\tkzGetPoint{C}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPoints(A,B,C)
 \tkzLabelPoints(A,B)
 \tkzLabelPoints[above](C)
 \tkzLabelAngle[pos=0.8](B,A,C){$72^\circ$}
 \tkzLabelAngle[pos=0.8](C,B,A){$72^\circ$}
 \tkzLabelAngle[pos=0.8](A,C,B){$36^\circ$}
\end{tikzpicture}
\end{tkzexample}

\section{Specific triangles with \tkzcname{tkzDefSpcTriangle}}

The centers of some triangles have been defined in the "points" section, here it is a question of determining the three vertices of specific triangles.

\begin{NewMacroBox}{tkzDefSpcTriangle}{\oarg{local options}\parg{p1,p2,p3}\marg{r1,r2,r3}}
The order of the points is important! p1p2p3 defines a triangle then the result is a triangle whose vertices have as reference a combination with \tkzname{name} and r1,r2, r3. If \tkzname{name} is empty then the references are  r1,r2 and r3.


\medskip
\begin{tabular}{lll}%
\toprule
options             & default & definition                        \\
\midrule
\TOline{orthic} {centroid}{determined by endpoints of the altitudes ...}
\TOline{centroid or medial}{centroid}{intersection of the triangle's three triangle medians}
\TOline{in or incentral}{centroid}{determined with the angle bisectors}
\TOline{ex or excentral} {centroid}{determined with the excenters}
\TOline{extouch}{centroid}{formed by the points of tangency  with the excircles}
\TOline{intouch or contact} {centroid}{formed by the points of tangency of the incircle}
\TOline{} {}{each of the vertices}
\TOline{euler} {centroid}{formed by Euler points on the  nine-point circle}
\TOline{symmedial} {centroid}{intersection points of the symmedians}
\TOline{tangential}{centroid}{formed by the lines tangent to the circumcircle}
\TOline{feuerbach} {centroid}{formed by the points of tangency of the nine-point ...}
\TOline{} {} {circle with the excircles}
\TOline{name} {empty}{used to name the vertices}
\midrule
\end{tabular}

\end{NewMacroBox}

\subsection{How to name the vertices}

With \tkzcname{tkzDefSpcTriangle[medial,name=M](A,B,C)\{\_A,\_B,\_C\}} you get three vertices named $M_A$, $M_B$ and $M_C$.

With \tkzcname{tkzDefSpcTriangle[medial](A,B,C)\{a,b,c\}} you get three vertices named and labeled $a$, $b$ and $c$.

Possible \tkzcname{tkzDefSpcTriangle[medial,name=M\_](A,B,C)\{A,B,C\}} you get three vertices named $M_A$, $M_B$ and $M_C$.

\subsection{Option \tkzname{medial} or \tkzname{centroid} }
The geometric centroid  of the polygon vertices of a triangle is the point $G$ (sometimes also denoted $M$) which is also the intersection of the triangle's three triangle medians. The point is therefore sometimes called the median point. The centroid is always in the interior of the triangle.
\\

\href{http://mathworld.wolfram.com/TriangleCentroid.html}{Weisstein, Eric W. "Centroid triangle" From MathWorld--A Wolfram Web Resource.}

In the following example, we obtain the Euler circle which passes through the previously defined points.

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[rotate=90,scale=.75]
 \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
 \tkzDefTriangleCenter[centroid](A,B,C)
 \tkzGetPoint{M}
 \tkzDefSpcTriangle[medial,name=M](A,B,C){_A,_B,_C}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawSegments[dashed,new](A,M_A B,M_B C,M_C)
 \tkzDrawPolygon[new](M_A,M_B,M_C)
 \tkzDrawPoints(A,B,C)
 \tkzDrawPoints[new](M,M_A,M_B,M_C)
 \tkzAutoLabelPoints[center=M,font=\scriptsize]%
(A,B,C,M_A,M_B,M_C)
 \tkzLabelPoints[font=\scriptsize](M)
\end{tikzpicture}
\end{tkzexample}

\subsection{Option \tkzname{in} or \tkzname{incentral} }

The incentral triangle is the triangle whose vertices are determined by
the intersections of the reference triangle’s angle bisectors with the
respective opposite sides.
\\
\href{http://mathworld.wolfram.com/ContactTriangle.html}{Weisstein, Eric W. "Incentral triangle" From MathWorld--A Wolfram Web Resource.}


\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=1]
  \tkzDefPoints{ 0/0/A,5/0/B,1/3/C}
  \tkzDefSpcTriangle[in,name=I](A,B,C){_a,_b,_c}
  \tkzInCenter(A,B,C)\tkzGetPoint{I}
  \tkzDrawPolygon(A,B,C)
  \tkzDrawPolygon[new](I_a,I_b,I_c)
  \tkzDrawPoints(A,B,C,I,I_a,I_b,I_c)
  \tkzDrawCircle[in](A,B,C)
  \tkzDrawSegments[dashed,new](A,I_a B,I_b C,I_c)
  \tkzAutoLabelPoints[center=I,%
  new,font=\scriptsize](I_a,I_b,I_c)
  \tkzAutoLabelPoints[center=I,
    font=\scriptsize](A,B,C)
\end{tikzpicture}
\end{tkzexample}

\subsection{Option \tkzname{ex} or \tkzname{excentral} }

The excentral triangle of a triangle $ABC$ is the triangle $J_aJ_bJ_c$ with vertices corresponding to the excenters of $ABC$.

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.6]
 \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
 \tkzDefSpcTriangle[excentral,name=J](A,B,C){_a,_b,_c}
 \tkzDefSpcTriangle[extouch,name=T](A,B,C){_a,_b,_c}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPolygon[new](J_a,J_b,J_c)
 \tkzClipBB
 \tkzDrawPoints(A,B,C)
 \tkzDrawPoints[new](J_a,J_b,J_c)
 \tkzLabelPoints(A,B,C)
 \tkzLabelPoints[new](J_b,J_c)
 \tkzLabelPoints[new,above](J_a)
 \tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c) 
\end{tikzpicture}
\end{tkzexample}


\subsection{Option \tkzname{intouch} or \tkzname{contact}}
The contact triangle of a triangle $ABC$, also called the intouch triangle, is the triangle  formed by the points of tangency of the incircle of $ABC$ with $ABC$.\\
\href{http://mathworld.wolfram.com/ContactTriangle.html}{Weisstein, Eric W. "Contact triangle" From MathWorld--A Wolfram Web Resource.}

We obtain the intersections of the bisectors with the sides.
\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
 \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}          
 \tkzDefSpcTriangle[intouch,name=X](A,B,C){_a,_b,_c}
 \tkzInCenter(A,B,C)\tkzGetPoint{I}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawPolygon[new](X_a,X_b,X_c)
 \tkzDrawPoints(A,B,C)
 \tkzDrawPoints[new](X_a,X_b,X_c)
 \tkzDrawCircle[in](A,B,C)
 \tkzAutoLabelPoints[center=I,blue,font=\scriptsize]%
(X_a,X_b,X_c)
 \tkzAutoLabelPoints[center=I,red,font=\scriptsize]%
(A,B,C)
\end{tikzpicture} 
\end{tkzexample}

\subsection{Option \tkzname{extouch}}
The extouch triangle  $T_aT_bT_c$ is the triangle formed by the points of tangency of a triangle $ABC$ with its excircles $J_a$, $J_b$, and $J_c$. The points  $T_a$, $T_b$, and $T_c$ can also be constructed as the points which bisect the perimeter of $A_1A_2A_3$ starting at $A$, $B$, and $C$.\\
\href{http://mathworld.wolfram.com/ExtouchTriangle.html}{Weisstein, Eric W. "Extouch triangle" From MathWorld--A Wolfram Web Resource.}

We obtain the points of contact of the exinscribed circles as well as the triangle formed by the centres of the exinscribed circles.

\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.7]
\tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
\tkzDefSpcTriangle[excentral,
                 name=J](A,B,C){_a,_b,_c}
\tkzDefSpcTriangle[extouch,
                  name=T](A,B,C){_a,_b,_c}
\tkzDefTriangleCenter[nagel](A,B,C)
\tkzGetPoint{N_a}
\tkzDefTriangleCenter[centroid](A,B,C)
\tkzGetPoint{G}
\tkzDrawPoints[new](J_a,J_b,J_c)
\tkzClipBB \tkzShowBB
\tkzDrawCircles[gray](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawLines[add=1 and 1](A,B B,C C,A)
\tkzDrawSegments[new](A,T_a B,T_b C,T_c)
\tkzDrawSegments[new](J_a,T_a J_b,T_b J_c,T_c)
\tkzDrawPolygon(A,B,C)
\tkzDrawPolygon[new](T_a,T_b,T_c)
\tkzDrawPoints(A,B,C,N_a)
\tkzLabelPoints(N_a)
\tkzAutoLabelPoints[center=N_a](A,B,C)
\tkzAutoLabelPoints[center=G,new,
                         dist=.4](T_a,T_b,T_c)
\tkzMarkRightAngles[fill=gray!15](J_a,T_a,B
 J_b,T_b,C J_c,T_c,A)
\end{tikzpicture}
\end{tkzexample}

\subsection{Option \tkzname{orthic}}

Given a triangle $ABC$, the triangle $H_AH_BH_C$ whose vertices are endpoints of the altitudes from each of the vertices of ABC is called the orthic triangle, or sometimes the altitude triangle. The three lines $AH_A$, $BH_B$, and $CH_C$ are concurrent at the orthocenter H of ABC.

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[scale=.75]
\tkzDefPoints{1/5/A,0/0/B,7/0/C}
 \tkzDefSpcTriangle[orthic](A,B,C){H_A,H_B,H_C}
 \tkzDefTriangleCenter[ortho](B,C,A)
 \tkzGetPoint{H}
 \tkzDefPointWith[orthogonal,normed](H_A,B)
 \tkzGetPoint{a}
 \tkzDrawSegments[new](A,H_A B,H_B C,H_C)   
 \tkzMarkRightAngles[fill=gray!20,
         opacity=.5](A,H_A,C B,H_B,A C,H_C,A)
 \tkzDrawPolygon[fill=teal!20,opacity=.3](A,B,C)
 \tkzDrawPoints(A,B,C)
 \tkzDrawPoints[new](H_A,H_B,H_C)
 \tkzDrawPolygon[new,fill=orange!20,
                opacity=.3](H_A,H_B,H_C)
 \tkzDrawPoint(a)
 \tkzLabelPoints(C)
 \tkzLabelPoints[left](B)
 \tkzLabelPoints[above](A)
 \tkzLabelPoints[new](H_A)
 \tkzLabelPoints[new,above left](H_C)
 \tkzLabelPoints[new,above right](H_B,H)
\end{tikzpicture}
\end{tkzexample}
    
\subsection{Option \tkzname{feuerbach}}
The Feuerbach triangle is the triangle formed by the three points of tangency of the nine-point circle with the excircles.\\
\href{http://mathworld.wolfram.com/FeuerbachTriangle.html}{Weisstein, Eric W. "Feuerbach triangle" From MathWorld--A Wolfram Web Resource.}

 The points of tangency define the Feuerbach triangle.

\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=1.25]
  \tkzDefPoint(0,0){A}
  \tkzDefPoint(3,0){B}
  \tkzDefPoint(0.5,2.5){C}
  \tkzDefCircle[euler](A,B,C) \tkzGetPoint{N}
  \tkzDefSpcTriangle[feuerbach,
                       name=F](A,B,C){_a,_b,_c}
  \tkzDefSpcTriangle[excentral,
                       name=J](A,B,C){_a,_b,_c}
  \tkzDefSpcTriangle[extouch,
                        name=T](A,B,C){_a,_b,_c}
  \tkzDrawPoints[blue](J_a,J_b,J_c,%
          F_a,F_b,F_c,A,B,C)
  \tkzClipBB \tkzShowBB
  \tkzDrawCircle[purple](N,F_a)
  \tkzDrawPolygon(A,B,C)
  \tkzDrawPolygon[new](F_a,F_b,F_c)
  \tkzDrawCircles[gray](J_a,F_a J_b,F_b J_c,F_c)
  \tkzAutoLabelPoints[center=N,dist=.3,
   font=\scriptsize](A,B,C,F_a,F_b,%
                   F_c,J_a,J_b,J_c)
\end{tikzpicture}
\end{tkzexample}

\subsection{Option   \tkzname{tangential}} 
The tangential triangle is the triangle $T_aT_bT_c$ formed by the lines tangent to the circumcircle of a given triangle $ABC$ at its vertices. It is therefore antipedal triangle of $ABC$ with respect to the circumcenter $O$.\\ 
\href{http://mathworld.wolfram.com/TangentialTriangle.html}{Weisstein, Eric W. "Tangential Triangle." From MathWorld--A Wolfram Web Resource. }


\begin{tkzexample}[latex=8cm,small]
\begin{tikzpicture}[scale=.5,rotate=80]
  \tkzDefPoints{0/0/A,6/0/B,1.8/4/C}           
  \tkzDefSpcTriangle[tangential,
    name=T](A,B,C){_a,_b,_c}
  \tkzDrawPolygon(A,B,C)
  \tkzDrawPolygon[new](T_a,T_b,T_c)
  \tkzDrawPoints(A,B,C)
  \tkzDrawPoints[new](T_a,T_b,T_c)
  \tkzDefCircle[circum](A,B,C)  
  \tkzGetPoint{O} 
  \tkzDrawCircle(O,A)
  \tkzLabelPoints(A,B,C)
  \tkzLabelPoints[new](T_a,T_b,T_c)
\end{tikzpicture} 
\end{tkzexample} 

\subsection{Option   \tkzname{euler}} 
The Euler triangle of a triangle $ABC$ is the triangle $E_AE_BE_C$ whose vertices are the midpoints of the segments joining the orthocenter $H$ with the respective vertices. The vertices of the triangle are known as the Euler points, and lie on the nine-point circle.
\\
\href{https://mathworld.wolfram.com/EulerTriangle.html}{Weisstein, Eric W. "Euler Triangle." From MathWorld--A Wolfram Web Resource.} 

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}[rotate=90,scale=1.25]
 \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
 \tkzDefSpcTriangle[medial,
     name=M](A,B,C){_A,_B,_C}
 \tkzDefTriangleCenter[euler](A,B,C)
     \tkzGetPoint{N} % I= N nine points
 \tkzDefTriangleCenter[ortho](A,B,C)
        \tkzGetPoint{H}
 \tkzDefMidPoint(A,H) \tkzGetPoint{E_A}
 \tkzDefMidPoint(C,H) \tkzGetPoint{E_C}
 \tkzDefMidPoint(B,H) \tkzGetPoint{E_B}
 \tkzDefSpcTriangle[ortho,name=H](A,B,C){_A,_B,_C}
 \tkzDrawPolygon(A,B,C)
 \tkzDrawCircle(N,E_A)
 \tkzDrawSegments[new](A,H_A B,H_B C,H_C)
 \tkzDrawPoints(A,B,C,N,H)
 \tkzDrawPoints[red](M_A,M_B,M_C)
 \tkzDrawPoints[blue]( H_A,H_B,H_C)
 \tkzDrawPoints[green](E_A,E_B,E_C)
 \tkzAutoLabelPoints[center=N,font=\scriptsize]%
(A,B,C,M_A,M_B,M_C,H_A,H_B,H_C,E_A,E_B,E_C)
\tkzLabelPoints[font=\scriptsize](H,N)
\tkzMarkSegments[mark=s|,size=3pt,
  color=blue,line width=1pt](B,E_B E_B,H)
   \tkzDrawPolygon[color=cyan](M_A,M_B,M_C)
\end{tikzpicture}
\end{tkzexample}

\subsection{Option  \tkzname{euler} and Option  \tkzname{orthic}} 
\begin{tkzexample}[vbox,small]
  \begin{tikzpicture}[scale=1.25]
    \tkzDefPoints{0/0/A,6/0/B,0.8/4/C}
    \tkzDefSpcTriangle[euler,name=E](A,B,C){a,b,c}
    \tkzDefSpcTriangle[orthic,name=H](A,B,C){a,b,c}
    \tkzDefExCircle(A,B,C) \tkzGetPoint{I} \tkzGetLength{rI}
    \tkzDefExCircle(C,A,B) \tkzGetPoint{J} \tkzGetLength{rJ}
    \tkzDefExCircle(B,C,A) \tkzGetPoint{K} \tkzGetLength{rK}
    \tkzDrawPoints[orange](I,J,K)
    \tkzLabelPoints[font=\scriptsize](A,B,C,I,J,K)
    \tkzClipBB
    \tkzInterLC[R](I,C)(I,\rI) \tkzGetSecondPoint{Fc}
    \tkzInterLC[R](J,B)(J,\rJ) \tkzGetSecondPoint{Fb}
    \tkzInterLC[R](K,A)(K,\rK) \tkzGetSecondPoint{Fa}
    \tkzDrawLines[add=1.5 and 1.5](A,B A,C B,C)
    \tkzDrawCircle[euler,orange](A,B,C) \tkzGetPoint{E}
    \tkzDrawSegments[orange](E,I E,J E,K)
    \tkzDrawSegments[dashed](A,Ha B,Hb C,Hc)
    \tkzDrawCircles[R](J,{\rJ} I,{\rI} K,{\rK})
    \tkzDrawPoints(A,B,C)
    \tkzDrawPoints[orange](E,I,J,K,Ha,Hb,Hc,Ea,Eb,Ec,Fa,Fb,Fc)
    \tkzLabelPoints[font=\scriptsize](E,Ea,Eb,Ec,Ha,Hb,Hc,Fa,Fb,Fc)  
  \end{tikzpicture}
\end{tkzexample}


\subsection{Option \tkzname{symmedial}}
The symmedial triangle$ K_AK_BK_C$ is the triangle whose vertices are the intersection points of the symmedians with the reference triangle $ABC$. 

\begin{tkzexample}[latex=7cm,small]
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(5,0){B}
\tkzDefPoint(.75,4){C}
\tkzDefTriangleCenter[symmedian](A,B,C)\tkzGetPoint{K} 
\tkzDefSpcTriangle[symmedial,name=K_](A,B,C){A,B,C}
\tkzDrawPolygon(A,B,C)
\tkzDrawSegments[new](A,K_A B,K_B C,K_C)
\tkzDrawPoints(A,B,C,K,K_A,K_B,K_C)
\tkzLabelPoints[font=\scriptsize](A,B,C,K,K_A,K_B,K_C)
\end{tikzpicture}
\end{tkzexample}
 
\endinput